3. The attempt at a solution
This will obviously involve a 2nd order diff eq, and there are enough initial conditions to solve for the unknown constants. If the potential were given with the r variable instead of x and y, it would be simpler. As such, I'm not sure how to take dV/dr when V is V(x,y) not V(r)...

You are doing it the wrong way, there are vector involved and ##\frac{d}{d \vec r} = \nabla ## and everyone already says it: force is the gradient of the potential, can you work it out now ?
[Edit: In case I wasn't clear, ##\nabla = \lt \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \gt ## ]